Difference between revisions of "Vieta's formulas"
Etmetalakret (talk | contribs) m |
Etmetalakret (talk | contribs) |
||
Line 13: | Line 13: | ||
When expanding this factorization of <math>P(x)</math>, each term is generated by a series of <math>n</math> choices of whether to include <math>x</math> or the negative root <math>-r_{i}</math> from every factor <math>(x-r_{i})</math>. Consider all the expanded terms of the polynomial with degree <math>n-j</math>; they are formed by multiplying a choice of <math>j</math> negative roots, making the remaining <math>n-j</math> choices in the product <math>x</math>, and finally multiplying by the constant <math>a_n</math>. | When expanding this factorization of <math>P(x)</math>, each term is generated by a series of <math>n</math> choices of whether to include <math>x</math> or the negative root <math>-r_{i}</math> from every factor <math>(x-r_{i})</math>. Consider all the expanded terms of the polynomial with degree <math>n-j</math>; they are formed by multiplying a choice of <math>j</math> negative roots, making the remaining <math>n-j</math> choices in the product <math>x</math>, and finally multiplying by the constant <math>a_n</math>. | ||
− | Note that adding together every multiplied choice of <math>j</math> negative roots yields <math>(-1)^j s_j</math>. Thus, when we expand <math>P(x)</math>, the coefficient of <math>x_{n-j}</math> is equal to <math>(-1)^j a_n s_j</math>. However, we defined the coefficient of <math>x^{n-j}</math> to be <math>a_{n-j}</math>. Thus, <math>(-1)^j a_n s_j = a_{n-j}</math>, or <math>s_j = (-1)^j a_{n-j}/a_n</math>, which completes the proof. <math>\ | + | Note that adding together every multiplied choice of <math>j</math> negative roots yields <math>(-1)^j s_j</math>. Thus, when we expand <math>P(x)</math>, the coefficient of <math>x_{n-j}</math> is equal to <math>(-1)^j a_n s_j</math>. However, we defined the coefficient of <math>x^{n-j}</math> to be <math>a_{n-j}</math>. Thus, <math>(-1)^j a_n s_j = a_{n-j}</math>, or <math>s_j = (-1)^j a_{n-j}/a_n</math>, which completes the proof. <math>\square</math> |
== Problems == | == Problems == |
Revision as of 11:08, 7 November 2021
In algebra, Vieta's formulas are a set of results that relate the coefficients of a polynomial to its roots. In particular, it states that the elementary symmetric polynomials of its roots can be easily expressed as a ratio between two of the polynomial's coefficients.
It is among the most ubiquitous results to circumvent finding a polynomial's roots in competition math and sees widespread usage in all mathematics contests.
Statement
Let be any polynomial with complex coefficients with roots
, and let
be the
elementary symmetric polynomial of the roots.
Vieta’s formulas then state that
This can be compactly summarized as
for some
such that
.
Proof
Let all terms be defined as above. By the factor theorem, . We will then prove Vieta’s formulas by expanding this polynomial and comparing the resulting coefficients with the original polynomial’s coefficients.
When expanding this factorization of , each term is generated by a series of
choices of whether to include
or the negative root
from every factor
. Consider all the expanded terms of the polynomial with degree
; they are formed by multiplying a choice of
negative roots, making the remaining
choices in the product
, and finally multiplying by the constant
.
Note that adding together every multiplied choice of negative roots yields
. Thus, when we expand
, the coefficient of
is equal to
. However, we defined the coefficient of
to be
. Thus,
, or
, which completes the proof.
Problems
Here are some problems with solutions that utilize Vieta's formulas.